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Canvas Example MCQ — Q91

• A) approximately -7.0 percentage points
• B) approximately -15.2 percentage points
• C) approximately -20.0 percentage points
• D) approximately +15.2 percentage points
• E) None of A-D is close; write the correct answer.

At MARGIN=2, z=.50 -> P=.6915. At MARGIN=4, z=.10 -> P=.5398. Change = .5398-.6915 = -.1517 = -15.17 pp.

Rule: For a discrete nonlinear change, calculate both probabilities and subtract.

Computer Lab 6 — Exercise 3, Task c

• A) ≈17.59%
• B) ≈82.41%
• C) ≈37.21%
• D) ≈93.12%

P = Φ(-0.9312) ≈ 0.1759 = 17.59%, reading the standard normal CDF at z = -0.9312.

Rule: Compute z=Xβ first, then look up Φ(z) for the predicted probability.

Computer Lab 6 — Exercise 3, Task c comparison

• A) +19.62 percentage points
• B) +211.6% relative change only, not expressible in percentage points
• C) -19.62 percentage points
• D) +1.962 percentage points

Change = 37.21% − 17.59% = 19.62 percentage points, consistent with the positive NETWORTH coefficient.

Rule: For discrete changes in nonlinear probability models, compute new probability minus old probability directly.

Computer Lab 6 — Exercise 3, Task f discrete

• A) -6.18 percentage points; Probit is nonlinear so the marginal effect changes as X changes, making it invalid to simply double the local marginal effect for a large discrete move
• B) -7.86 percentage points (exactly 2×-3.93), because Probit effects are linear
• C) +6.18 percentage points, since probability must rise as MARGIN rises
• D) 0 percentage points, since probability is bounded between 0 and 1

Change = 5.99% − 12.16% = -6.18 percentage points. Because Probit is nonlinear, φ(Xβ) differs at MARGIN=2 versus MARGIN=4, so simply doubling the local marginal effect (which assumes linearity) would be inaccurate; always compute P_new − P_old for discrete changes.

Rule: For discrete changes in nonlinear probability models, always use P(new X) − P(old X), not linear-effect×Δx.

Computer Lab 6 — Exercise 4, LPM weaknesses

• A) Predicted probabilities can fall outside [0,1], and errors are heteroskedastic by construction, so robust standard errors are generally preferred
• B) LPM cannot estimate coefficients at all for binary outcomes
• C) LPM coefficients cannot be interpreted in percentage-point terms
• D) LPM always has a lower R² than Probit or Logit

Because LPM is linear, predicted probabilities are not constrained to [0,1], and the binary nature of Y produces heteroskedastic errors, making robust standard errors advisable, even though the coefficients remain directly interpretable in probability terms.

Rule: Key LPM limitations: possible out-of-range probabilities and heteroskedastic errors; use robust SEs.

Computer Lab 6 — Exercise 4, Probit vs LPM nonlinearity

• A) Because Probit is nonlinear: the marginal/discrete effect of PIRAT depends on the predicted probability level itself, which differs by BLACK status even without an explicit interaction term
• B) Because the Probit model actually includes an interaction term between PIRAT and BLACK
• C) Because Probit coefficients differ by group automatically
• D) This is a calculation error; Probit should give identical effects just like LPM

Even without an interaction term, Probit's nonlinear Φ(·) link function makes the implied probability effect of any regressor depend on the overall linear index (and thus on BLACK, which shifts the index), unlike LPM where effects are mechanically constant across groups.

Rule: Nonlinear models (Probit/Logit) naturally produce covariate-dependent marginal effects even without explicit interaction terms, unlike the linear LPM.

Computer Lab 6 — Exercise 4, Logit calc

• A) ≈27.28%
• B) ≈72.72%
• C) ≈49.51%
• D) ≈9.05%

Λ(-0.9805) = 1/(1+e^0.9805) = 1/(1+2.6654) ≈ 1/3.6654 ≈ 0.2728 = 27.28%, matching the lab's reported value.

Rule: Logit predicted probability: Λ(z) = 1/(1+e^(-z)), computed directly from the linear index z.